GRE Math: Units & Remainders (Level 2)

Solution (A): 1. For \(8^{102}\): Cycle is {8, 4, 2, 6}. \(102 \div 4 = 25\) remainder 2. This is the 2nd step: 4. 2. For \(3^{63}\): Cycle is {3, 9, 7, 1}. \(63 \div 4 = 15\) remainder 3. This is the 3rd step: 7. 3. Sum of units digits: \(4 + 7 = 11\). The final units digit is 1.

Solution (D): 1. For \((144)^{44}\): Base ends in 4. Cycle is {4, 6}. The exponent 44 is even, so it’s the 2nd step: 6. 2. For \((177)^{77}\): Base ends in 7. Cycle is {7, 9, 3, 1}. \(77 \div 4 = 19\) remainder 1. This is the 1st step: 7. 3. Product of units digits: \(6 \times 7 = 42\). The final units digit is 2. *Correction:* Let me re-check. 4-even is 6. 7-rem1 is 7. 6*7 = 42 -> 2. The answer should be A.

Solution (A): 1. For \((144)^{44}\): We only care about \(4^{44}\). The cycle for 4 is {4, 6}. Since the exponent 44 is even, the units digit is 6. 2. For \((177)^{77}\): We only care about \(7^{77}\). The cycle for 7 is {7, 9, 3, 1}. \(77 \div 4 = 19\) remainder 1. This corresponds to the 1st step: 7. 3. Product of units digits: \(6 \times 7 = 42\). The final units digit is 2.

Solution (A): The remainder when divided by 5 is determined by the units digit. We need the units digit of \(13^{100}\). We look at \(3^{100}\). The cycle for 3 is {3, 9, 7, 1}. The exponent 100 is divisible by 4 (remainder 0), so it’s the 4th step: 1. The units digit is 1. Any number ending in 1, when divided by 5, has a remainder of 1.

Solution (E): 1. For \(7^{4n+2}\): The exponent \(4n+2\) when divided by 4 always has a remainder of 2. The 2nd step in the 7-cycle {7, 9, 3, 1} is 9. 2. For \(9^{2n+1}\): The exponent \(2n+1\) is always odd. The cycle for 9 is {9, 1}. An odd exponent corresponds to the 1st step: 9. 3. Sum of units digits: \(9 + 9 = 18\). The final units digit is 8.

Solution (C): First, evaluate the exponent: \(3^4 = 81\). The problem is now to find the units digit of \(7^{81}\). The cycle for 7 is {7, 9, 3, 1}. We need the remainder of the exponent 81 when divided by 4. \(81 \div 4 = 20\) with a remainder of 1. A remainder of 1 corresponds to the 1st step in the cycle. The units digit is 7.

Solution (B): The remainder when divided by 10 is the units digit of the sum \(S\). \(2^1 = 2\) \(2^2 = 4\) \(2^3 = 8\) \(2^4 = 16 \to\) units digit 6 \(2^5 = 32 \to\) units digit 2 Sum of units digits: \(2 + 4 + 8 + 6 + 2 = 22\). The units digit of the sum is 2. Therefore, the remainder is 2.

Solution (A): We can use remainders (modular arithmetic). \(3^1 = 3 \equiv -1 \pmod{4}\) \(3^2 = 9 \equiv 1 \pmod{4}\) We can write \(3^{100}\) as \((3^2)^{50}\). This is equivalent to \((1)^{50} \pmod{4}\), which is just 1. Alternatively, \(3^{100}\) is an odd power of 9. Powers of 9 always end in 81 or 09, etc. Let’s use the first method. \(3^{100} = (3^2)^{50} = 9^{50}\). \(9 \div 4 = 2\) remainder 1. So \(9 \equiv 1 \pmod{4}\). \(9^{50} \equiv 1^{50} \pmod{4} \equiv 1\). The remainder is 1.

Solution (C): We only need the units digits of each part. 1. For \((12)^3\): We need \(2^3\). Cycle {2, 4, 8, 6}. 3rd step is 8. 2. For \((13)^4\): We need \(3^4\). Cycle {3, 9, 7, 1}. 4th step is 1. 3. For \((14)^5\): We need \(4^5\). Cycle {4, 6}. Exponent 5 is odd, so 1st step is 4. 4. Product of units digits: \(8 \times 1 \times 4 = 32\). The final units digit is 2. *Correction:* Let’s re-check. 2^3=8. 3^4=81 -> 1. 4^5 (odd) -> 4. 8*1*4 = 32 -> 2. The answer should be B.

Solution (B): We only need the units digits of each part. 1. For \((12)^3\): We need the units digit of \(2^3\), which is 8. 2. For \((13)^4\): We need the units digit of \(3^4\). The cycle is {3, 9, 7, 1}. The 4th step is 1. 3. For \((14)^5\): We need the units digit of \(4^5\). The cycle is {4, 6}. Since the exponent 5 is odd, the units digit is 4. 4. Product of units digits: \(8 \times 1 \times 4 = 32\). The final units digit is 2.

Solution (D): If the units digit of \(n^2\) is 1, the units digit of \(n\) must be either 1 (since \(1^2=1\)) or 9 (since \(9^2=81\)). Case 1: Units digit of \(n\) is 1. Then the units digit of \((n+1)\) is \(1+1 = 2\). The units digit of \((n+1)^2\) is \(2^2 = 4\). Case 2: Units digit of \(n\) is 9. Then the units digit of \((n+1)\) is \(9+1 = 10 \to 0\). The units digit of \((n+1)^2\) is \(0^2 = 0\). The possible units digits are 4 and 0. Of the options provided, 4 is a possibility.

Solution (A): The remainder when divided by 10 is the units digit. 1. For \(5^{1000}\): Any power of 5 (after \(5^1\)) has a units digit of 5. 2. For \(6^{1000}\): Any power of 6 has a units digit of 6. 3. Sum of units digits: \(5 + 6 = 11\). The units digit of the sum is 1. Therefore, the remainder is 1.